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Sample Proofs
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\noindent \textbf{Theorem.}
\textit{There are infinitely many prime numbers.}

\noindent \textit{Proof}. 
Suppose, towards a contradiction, that there are not infinitely many prime numbers. Then there are only finitely many, so we can list them as $p_1,p_2,\ldots,p_k$, for some integer $k\geq 1$. Define 
$$
N=p_1\cdot p_2\cdot\ldots\cdot p_k.
$$
Then the integer $N+1$ is at least 2, and so must have some prime factor $q$. By assumption, $q$ divides $N+1$. By definition, $q$ divides $N$, since it must be one of the primes $p_1,\ldots,p_k$. Therefore $q$ divides $N+1-N=1$. But this is a contradiction, since $q$ is prime and 1 is not divisible by any prime number.\qed

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\noindent\textbf{Theorem.}
\textit{$\ln 2$ is irrational.}

\noindent \textit{Proof}. 
Suppose, towards a contradiction, that $\ln 2$ is rational. Then $\ln 2$ can be written as $\frac{m}{n}$, where $m,n\in\mathbb{Z}$, $n\neq 0$, and $m\geq 0$. Note that $m\neq 0$, since otherwise we would have $\ln 2=0$, which is not true. So we may assume $m>0$. Then
\begin{align*}
\ln 2=\frac{m}{n} &\Imp 2=e^{\frac{m}{n}}\\
 &\Imp 2^n=e^m.
\end{align*}
Therefore $e$ is a root of the polynomial $x^m-2^n$, which implies that $e$ is an algebraic number. This is a contradiction, since $e$ is known to be transcendental (not algebraic).\qed

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